Why is the Fermi level (energy) shfited in doped semiconductors?

Question

According to

the Fermi level is shifted due to doping: Upwards in case of n-type and downwards in case of p-type.

Why is this?

The Fermi level is the level where the probability that an electron occupies the state is $0.5$, e.g. where the Fermi energy is located (correct?). I can understand that the distribution changes with the temperatures (it gets broader) but I don't understand why/how the Fermi level changes.

The Fermi energy is defined as:

$$ E_F = \frac{h^2}{2m_0} \frac{3 \pi^2 N}{V}^{3/2} $$

I can't see any “doping influence” in this equation. What moves the Fermi energy, and thus the Fermi level?

There are related questions like Why does Fermi Level change due to change in donor atom concentration? and Does the Fermi level depend on temperature? but they do not answer this question here.

Answer 1

Firstly , I should say a little of semiconductor background is needed to understand the shift in Fermi level. The following things might help you in this regards:

1. The thermal equilibrium carrier concentrations expressions: $n=N_{c}exp(-\frac{(E-E_{f})}{kT})$ and $p=N_{v}exp(-\frac{E_{f}-E_{v}}{kT})$
2. One must get the $E_{f}=\frac{(E_{c}-E_{v})}{2}$ from a condition that for an intrinsic semiconductor $n=p$ and also we assume that effective masses are same in both bands that is $m_{c}^{\ast}=m_{v}^{\ast}$. As $E_{g}=E_{C}-E_{v}$ bandgap therefore we obtain $E_{f}=\frac{E_{g}}{2} $ or $E_{f}=E_{c}-\frac{E_{g}}{2}$ means that the Fermi level is right in the middle of the bandgap for intrinsic semicondusctors.
3. Now, let us consider we doped the intrinsic semiconductor so that it became N-type semiconductor which in turn means that the majority carrier concentrations is now $n=N_{d}$ , where $N_{d}$ stands for donors concentrations. After certain manipulation with expressions oneust endup with the following new expression for the New Fermi level($E_{F}$) in terms of the intrisic fermi level $E_{f}$: $E_{F}=E_{f}+k_{B}Tln(\frac{N_{v}}{N_{c}}){(\frac{N_{d}}{n_i})}^2$ where $N_{d}\gt n_{i}$.Now We can see that the new Fermi Level is above the previous one. Thus it has shifted towards the conduction band.In case of P-type semiconductor the same procedure can be apllied to show that Fermi level will shift towrds the Valence band.

Answer 2

The key point is the fact that, while in the intrinsic material the electronic states are organized in a valence band and in a conduction band separated by a gap, in a n-type material (for example) there are more electronic states lying in the middle of the intrinsic gap, typycally very close to the conduction band. These electronic states appear as a consequence of introducing impurities in the lattice structure of the material. In a n-type material, the total number of electrons is also larger than in the intrinsic case, because the impurities have more valence electrons than the ions of the intrinsic lattice, so at zero temperature, not only the valence band, but also all these new electronic states are fully populated.

Since the gap between these "new" states and the conduction band is much smaller than the gap between $E_V$ and $E_C$, at a given temperature the fraction of electrons populating the conduction band is much larger than in the intrinsic material, and the Fermi function is modified as shown in the central panel of the figure. In turn, the energy at which the Fermi function is $1/2$ (the Fermi level!) is closer to $E_C$ and further from $E_V$, while in the intrinsic material it is right in the middle, due to the particle-hole symmetry.

One more thing: I would not use the formula $E_F \propto (N/V)^{2/3}$ you have mentioned in this context: this is the Fermi level for a gas of free fermions at zero temperature, while here the electrons are subject to the periodic potential of the lattice.

Hope this helps :)